Question

The ratio of the $2p → 1s$ transition energy in $He^+$ to that in the H atom is closest to

• A. 1
• B. 2
• C. 4
• D. 8

Answer 1

The Correct Option is C

To find the ratio of the transition energy for the $2p \rightarrow 1s$ transition in \(He^+\) to that in the H atom, we need to consider the Rydberg formula for calculating the energy of the \( nth \) level in hydrogen-like atoms. The formula for the energy levels is given by:

\(E_n = -Z^2 \frac{R_H}{n^2}\)

where:

• \(E_n\) is the energy of the level.
• \(Z\) is the atomic number of the element.
• \(R_H\) is the Rydberg constant for the hydrogen atom.
• \(n\) is the principal quantum number.

For a transition from the \(2p\) level (\(n=2\)) to the \(1s\) level (\(n=1\)), the energy change is given by:

\(\Delta E = E_1 - E_2 = -Z^2 \frac{R_H}{1^2} + Z^2 \frac{R_H}{2^2}\)

Simplifying this expression gives:

\(\Delta E = -Z^2 R_H + \frac{Z^2 R_H}{4} = -\frac{3Z^2 R_H}{4}\)

Now, let's calculate the transition energy ratio between \(He^+\) and \(H\):

• For \(H\), \(Z = 1\), so \(\Delta E_H = -\frac{3(1)^2 R_H}{4} = -\frac{3R_H}{4}\).
• For \(He^+\), \(Z = 2\), so \(\Delta E_{He^+} = -\frac{3(2)^2 R_H}{4} = -\frac{12R_H}{4} = -3R_H\).

The ratio of the transition energies is:

\(\text{Ratio} = \frac{\Delta E_{He^+}}{\Delta E_{H}} = \frac{-3R_H}{-\frac{3R_H}{4}} = 4\)

Therefore, the correct answer is 4. This shows that the energy of the \(2p \rightarrow 1s\) transition in \(He^+\) is four times that in the hydrogen atom.